MATH SOLVE

4 months ago

Q:
# Each base angle is an isosceles triangle measures 55 degrees 30 minutes. Each of the congruent sides is 10 centimeters long. Estimate the following problems to the nearest tenth. A. Find the altitude of the triangle. B.what is the length of the base? C.fund the area of the triangle.

Accepted Solution

A:

Answers:

55 deg 30 Minutes = 55.5 degrees

sin = opp/hyp

sin(55.5) = altitude/10

Altitude = 10*sin(55.5)

Altitude = 10 * sin(55.5) = 8.241261886

Rounded to nearest 10th

Altitude = 8.2 cm

b. for base use Low of Cosines

Let C = base

Side A and B = length 10 Cm

c^2 = 2*a^2 -2a^2*cos(<C)

<C = 180 - 2*55.5 = 180 -111 = 69 degrees

c = sqrt ( 2a^2 - 2a^2*cos(<C)

c = sqrt ( 2*100- 200*cos(69))

c = sqrt (200 - 200*cos(69) ) = 11.32812474

Answer base

c = 11.3 cms

c.

Area = (1/2)*b * h

Area = (1/2)*(answer a * answer b)

Area = (1/2)* 11.32812474* 8.241261886 = 46.67902132

Area = 46.7 (rounded)

Area = ( 1/2) * a*b*sin(<C) = (1/2)*10*10*sin(69) =

Area = 50* sin(69) = 46.67902132

Area = 46.7 (rounded)

55 deg 30 Minutes = 55.5 degrees

sin = opp/hyp

sin(55.5) = altitude/10

Altitude = 10*sin(55.5)

Altitude = 10 * sin(55.5) = 8.241261886

Rounded to nearest 10th

Altitude = 8.2 cm

b. for base use Low of Cosines

Let C = base

Side A and B = length 10 Cm

c^2 = 2*a^2 -2a^2*cos(<C)

<C = 180 - 2*55.5 = 180 -111 = 69 degrees

c = sqrt ( 2a^2 - 2a^2*cos(<C)

c = sqrt ( 2*100- 200*cos(69))

c = sqrt (200 - 200*cos(69) ) = 11.32812474

Answer base

c = 11.3 cms

c.

Area = (1/2)*b * h

Area = (1/2)*(answer a * answer b)

Area = (1/2)* 11.32812474* 8.241261886 = 46.67902132

Area = 46.7 (rounded)

Area = ( 1/2) * a*b*sin(<C) = (1/2)*10*10*sin(69) =

Area = 50* sin(69) = 46.67902132

Area = 46.7 (rounded)